- A$\frac{\pi }{4}$
- B$sin^{-1}\sqrt {\frac{2}{3}} $
- ✓$cot^{-1} \sqrt {2} $
- D$\frac{\pi }{3}$
$x^2 + h^2 = a^2$
$V = \frac{1}{3}y^2 h = \frac{1}{3} 2 x^2 h$ (note $4x^2 = 2y^2 ==> y^2 = 2x^2$)
$V (\alpha ) = \frac{2}{3}a^2 cos^2 \alpha . a sin \alpha = \frac{2}{3}a^3 sin \alpha cos^2 \alpha $
now $ V' (\alpha ) = 0$ ==> $tan \alpha = \frac{1}{{\sqrt 2 }}$ ;
$V_{max} =\frac{{4\,\sqrt 3 \,\,{a^3}}}{{27}}$ 
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$(\lambda-1) x+(3 \lambda+1) y+2 \lambda z=0$
$(\lambda-1) x+(4 \lambda-2) y+(\lambda+3) z=0$
$2 x+(3 \lambda+1) y+3(\lambda-1) z=0$
has non-zero solutions, is
$1.$ The correct statement$(s)$ is(are)
$(A)$ $f^{\prime}(1) < 0$
$(B)$ $f(2) < 0$
$(C)$ $f^{\prime}(x) \neq 0$ for any $x \in(1,3)$
$(D)$ $f^{\prime}(x)=0$ for some $x \in(1,3)$
$2.$ If $\int_1^3 x^2 F^{\prime}(x) d x=-12$ and $\int_1^3 x^3 F^{\prime \prime}(x) d x=40$, then the correct expression$(s)$ is(are)
$(A)$ $9 f^{\prime}(3)+f^{\prime}(1)-32=0$
$(B)$ $\int_1^3 f(x) d x=12$
$(C)$ $9 f^{\prime}(3)-f^{\prime}(1)+32=0$
$(D)$ $\int_1^3 f(x) d x=-12$
Give the answer question $1$ and $2.$